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Boltzmann equation for photons

  1. Jun 30, 2014 #1

    WannabeNewton

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    Hi there. I have a couple of questions regarding the derivation of the Boltzmann equation in Dodelson for photons when given scalar overdensity perturbations to the FRW metric.

    To start with, let ##\Theta(\vec{x})## denote the temperature perturbations to the Bose-Einstein distribution of the CMB gas and let ##\Theta(\vec{x}) = \int \frac{d^3 k}{(2\pi)^3}e^{i\vec{k}\cdot \vec{x}}\Theta(\vec{k})## denote its Fourier decomposition. Dodelson states, in the second paragraph of p.101, that "The wavevector ##\vec{k}## is pointing in the direction in which the temperature is changing, so it is perpendicular to the gradient..." but I'm not completely sure as to what he means by this.

    Is he simply referring to each ##\vec{k}## being perpendicular to the gradient of the associated Fourier modes given by ##\Theta_{\vec{k}}(\vec{x}) = \Theta(\vec{k})e^{i\vec{k}\cdot \vec{x}}##? But then ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x}) = i\vec{k}\Theta_{\vec{k}}(\vec{x})## so each Fourier mode is such that ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})## is parallel to ##\vec{k}##, not perpendicular to it. In fact for each mode ##\vec{k}##, ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})## itself points in the direction in which the temperature changes so, by Dodelson's own remark "The wavevector ##\vec{k}## is pointing in the direction in which the temperature is changing", it must be that ##\vec{k}## is parallel to ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})##; his second remark "...so it is perpendicular to the gradient..." makes no sense in light of that and in fact it would be the level curves of ##\Theta_{\vec{k}}(\vec{x})## that ##\vec{k}## is perpendicular to.

    My second question is with regards to his statement atop p.99 that "In the absence of a bulk velocity for the electrons (##v_b = 0##), the [Compton] collision terms serve to drive ##\Theta## to ##\Theta_0##". Here ##\Theta_0## is the monopole moment of ##\Theta##. But if we already have ##v_b = 0## for the average velocity of the electron fluid then why would there be e.g. a dipole moment ##\Theta_1## of ##\Theta## to be driven to ##\Theta_0## under efficient Compton scattering? If ##v_b = 0## then wouldn't we already have only a monopole distribution for the temperature perturbations?

    The phrase "...the [Compton] collision terms serve to drive ##\Theta## to ##\Theta_0##" makes it seem like there is, amongst higher moments, an existing dipole moment even when ##v_b = 0## and that eventually it gets suppressed by the monopole moment after many Compton scatterings but if ##v_b = 0## then how can there be a preferred direction for there to even be a dipole moment to be suppressed to start with? Wouldn't it make more sense to say that ##v_b \neq 0## and ##\Theta_1 \neq 0## at first but that efficient Compton scatterings serve to drive both ##\Theta \rightarrow \Theta_0## and ##v_b \rightarrow 0##? Or is he saying that inflation can, in general, produce temperature perturbations for which the dipole moment ##\Theta_1 \neq 0##, irrespective of the average fluid velocity of the (non-relativistic) electrons which only come in later after the inflaton decays during reheating, and that after inflation and reheating are over and photons start scattering off electrons, if ##v_b = 0## then the ##\Theta_1## and possibly higher moments generated by inflation get suppressed by the monopole moment?

    Thank you in advance!
     
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  3. Jun 30, 2014 #2

    PeterDonis

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    In the usual 4-vector formalism, since the wave vector is null, it is orthogonal to itself, so any vector parallel to it is also orthogonal to it. One of the fun counterintuitive properties of null vectors. :wink:

    There is also a 2-dimensional space of spacelike vectors that are orthogonal to the wave vector (as is true of any null vector), and the level curves would lie in this subspace, so they will also be orthogonal to the wave vector, yes.

    However, it's not clear to me whether the vectors being used are supposed to be 4-vectors or 3-vectors. If he's using 3-vectors but interpreting "perpendicular" as I have above, that seems a bit weird.
     
  4. Jun 30, 2014 #3

    George Jones

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    What you wrote seemed reasonable, and the first thing I did was to "pick out" your modes by letting ##\Theta \left(\vec{k}\right) = \delta \left( \vec{k} - \vec{k_0}\right)##.

    Then I thought "Check the errata."

    http://home.fnal.gov/~dodelson/errata.html
     
  5. Jun 30, 2014 #4

    George Jones

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    I have had a brief look at page 99. With respect to your second question, isn't the first sentence on page 99 key?

    "Already, we can anticipate the effect of Compton scattering on the photon distribution."

    Doesn't this mean that ##\Theta## is a perturbation of the photon distribution? These perturbations to the photon distribution can have non-zero multipoles, but these are driven to zero by scattering off electrons that have zero bulk velocity (electron monopole distribution).
     
  6. Jun 30, 2014 #5

    WannabeNewton

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    They're supposed to be 3-vectors and the notions of parallel and perpendicular are supposed to be with respect to a Euclidean metric so I don't think the pseudo-Riemannian notion of a null vector applies.

    Ah! Thank you for that. It certainly clears things up haha.

    So just to clarify, the dipole and higher multipoles of ##\Theta## are generated during inflation and are already present before Compton scattering starts taking place and it is only after this that the Compton scattering drives ##\Theta## to ##\Theta_0## if ##v_b = 0##? In other words, I'm just wondering what causes the dipole and higher multipoles of the temperature perturbation ##\Theta## to come into existence in the first place, if not from inflation.

    Also just as a quick aside, on p.101 when Dodelson writes "We will typically assume that the velocity points in the same direction as ##\vec{k}## (this is equivalent to saying that the velocity is irrotational)"...I can see that ##\vec{v}_b \propto \vec{k} \propto \vec{\nabla}\Theta_{\vec{k}}## implies ##\vec{\nabla}\times \vec{v}_b \propto \vec{\nabla}\times\vec{\nabla}\Theta_{\vec{k}} = 0## however conversely if ##\vec{\nabla}\times \vec{v}_b = 0## then ##\vec{v}_b \propto \vec{\nabla}f## for some function ##f## but why should ##\vec{\nabla}f\propto \vec{k}##? This would have to hold if the velocity field of the electron fluid being parallel to the wave-vector of each Fourier mode is to be equivalent to being irrotational.

    Thanks to both of you!
     
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